Question

(a) Find the height from the earth's surface where $g$ will be $25\%$ of its value on the surface of earth.

($R=6400Km$).
(b) Find the percentage decrease in the value of $g$ at a depth $h$ from the surface of earth.

Answer 1

(a)At earth's surface $g=\frac{GM}{R^2}$..(1)

At x,$g^{\prime }=0.25g=\frac{g}{4}=\frac{GM}{x^2}$ ..(2)

(1) $÷$ (2)

$4=\frac{x^2}{R^{@}}$

$x=2R$

$x=2\times 6400=12800km$

Height from earth's surface is $(x-R)=6400km$

(b)At surface, $g=\frac{GM}{R^2}=\frac{G}{R^2}\times \rho \times \frac{4}{3}\pi R^3$

$=\frac{4}{3}\rho G\pi R$

At depth h, $\frac{g^{\prime }}{g}=\frac{R-h}{R}=1-\frac{h}{R}$

$g^{\prime }=g(1-\frac{h}{R})$

% decrease, $\frac{g-g^{\prime }}{g}=\frac{g-\frac{h}{R}}{g(1-\frac{h}{R})}=\frac{h}{R-h}$

%decrease=$\frac{100h}{R-h}$